//Given a binary tree, find the lowest common ancestor (LCA) of two given nodes 
//in the tree. 
//
// According to the definition of LCA on Wikipedia: “The lowest common ancestor 
//is defined between two nodes p and q as the lowest node in T that has both p and
// q as descendants (where we allow a node to be a descendant of itself).” 
//
// 
// Example 1: 
//
// 
//Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
//Output: 3
//Explanation: The LCA of nodes 5 and 1 is 3.
// 
//
// Example 2: 
//
// 
//Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
//Output: 5
//Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant o
//f itself according to the LCA definition.
// 
//
// Example 3: 
//
// 
//Input: root = [1,2], p = 1, q = 2
//Output: 1
// 
//
// 
// Constraints: 
//
// 
// The number of nodes in the tree is in the range [2, 105]. 
// -109 <= Node.val <= 109 
// All Node.val are unique. 
// p != q 
// p and q will exist in the tree. 
// 
// Related Topics 树 
// 👍 858 👎 0


//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode result;
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        if (root == p || root == q) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left == null) {
            return right;
        }
        if (right == null) {
            return left;
        }
        return root;

//        dfs(root,p,q);
//        return result;
    }
    public boolean dfs(TreeNode root,TreeNode p,TreeNode q) {
        if(root == null) {
            return false;
        }
        boolean lson = dfs(root.left,p,q);
        boolean rson = dfs(root.right,p,q);
        if(lson && rson || ((root == p || root == q) && (lson || rson))){
            result = root;
        }
        return lson || rson || root == p || root == q;
    }
}
//leetcode submit region end(Prohibit modification and deletion)
